Column 650x450 in double curvature

Data and assumptions

• Column 450 mm (width $b$) × 650 mm (depth $D$); $P_u = 1600$ kN; $f_{ck} = 30$ MPa, $f_y = 500$ MPa. Clear cover 40 mm (IS 456 cl. 26.4.2.1), 8 mm ties.
• Moments: major-axis (x-x) moments 90 kN·m (top) and 35 kN·m (bottom); minor-axis (y-y) moments 55 kN·m (top) and 30 kN·m (bottom). The column bends in double curvature, so the smaller end moment is taken as negative.
• The unsupported length for the minimum eccentricity is taken equal to the larger effective length, 7 m.
• Sizes are given as $b \times D$ with $D \ge b$; moments $M_{ux}$ act about the major axis (depth $D$) and $M_{uy}$ about the minor axis (depth $b$).

Slenderness

Minimum eccentricity (cl. 25.4)

$$e_{min} = \max\left(\frac{L}{500} + \frac{D}{30},\ 20\right)$$

• About x-x: $e_x = 7000/500 + 650/30 = 35.7$ mm (min. 20 mm); adopt 35.7 mm; $M_{ux,min} = P_u e = 57.1$ kN·m
• About y-y: $e_y = 7000/500 + 450/30 = 29.0$ mm (min. 20 mm); adopt 29.0 mm; $M_{uy,min} = 46.4$ kN·m

Design moments

• Major axis (short about x-x): no additional moment; the design moment is the larger end moment, $M_{ux} = 90.0$ kN·m ($> P_u e_{x,min} = 57.1$ kN·m).
• Minor axis (slender): $M_i = 0.6M_2 + 0.4M_1 = 0.6(55) + 0.4(-30) = 21.0$ kN·m, not less than $0.4M_2 = 22.0$; adopt 22.0 kN·m (cl. 39.7.1). Additional moment due to slenderness: $M_a = \dfrac{P_u b}{2000}\left(\dfrac{l_{ey}}{b}\right)^2 = \dfrac{1600 \times 0.45}{2000} \times 14.44^2 = 75.1$ kN·m, reduced by $k = \dfrac{P_{uz}-P_u}{P_{uz}-P_b} \le 1$.
• For the selected steel: $P_{uz} = 4821$ kN; $P_{by} = 1107$ kN, $k_y = 0.87$, $M_{ay} = 65.1$ kN·m.
• Design moments: $M_{ux} = 90.0$ kN·m, $M_{uy} = 22.0 + 65.1 = 87.1$ kN·m (not less than $M_2 = 55$ kN·m or the minimum-eccentricity values).

Longitudinal reinforcement

$\dfrac{P_u}{f_{ck} b D} = \dfrac{1600000}{30 \times 450 \times 650} = 0.182$ Bars are chosen so that $p$ lies between 0.8% (minimum) and 4% (practical maximum; IS 456 cl. 26.5.3.1 allows 6%). The uniaxial capacities $M_{ux1}$, $M_{uy1}$ at $P_u$ are read from the SP-16 interaction charts for rectangular sections; here they are computed by strain compatibility with the IS 456 stress block and the design stress-strain curve of steel (same basis as the charts), with effective cover $d' = 40 + 8 + \phi/2$.

Check

$P_{uz} = 0.45 f_{ck} A_c + 0.75 f_y A_{sc} = 0.45 \times 30 \times 290087 + 0.75 \times 500 \times 2413 = 4821$ kN $\dfrac{P_u}{P_{uz}} = \dfrac{1600}{4821} = 0.332$, so $\alpha_n = 1.22$ (IS 456 cl. 39.6; linear between 1.0 at 0.2 and 2.0 at 0.8).

$$\left(\frac{M_{ux}}{M_{ux1}}\right)^{\alpha_n} + \left(\frac{M_{uy}}{M_{uy1}}\right)^{\alpha_n} = \left(\frac{90.0}{492.3}\right)^{1.22} + \left(\frac{87.1}{325.0}\right)^{1.22} = 0.126 + 0.201 = 0.326 \le 1.0$$

The interaction condition is satisfied; the section is safe.

Provide 12-16 mm bars ($A_{sc} = 2413$ mm², $p = 0.82\%$), within 0.8% to 4%.

Lateral ties

• Diameter $\ge \max(\phi_{max}/4,\ 6) = \max(16/4, 6) = 6$ mm → provide 8 mm.
• Pitch $\le \min$(least lateral dimension, $16\phi_{min}$, 300 mm) $= \min(450, 256, 300) = 256$ mm → provide 8 mm ties @ 250 mm c/c.

Detailing

CROSS-SECTION 450 x 650
    +-------------------------+
    |  o      o     o      o  |
    |                         |
    |                         |
    |  o                   o  |
    |                         |
    |                         |
    |                         |
    |  o                   o  |
    |                         |
    |                         |
    |  o      o     o      o  |
    +-------------------------+
   o = 16 mm bars, ties 8 mm @ 250 c/c

Answer: 450 × 650 mm column with 12-16 mm longitudinal bars ($p = 0.82\%$) and 8 mm ties @ 250 mm c/c.